Approximation Algorithms: Combinatorial Approaches

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Approximation Algorithms Combinatorial Approaches

• Lecture 13 March 2

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The Hamiltonian Path Problem
Hamiltonian Path Problem Given an undirected
graph, find a cycle visiting every vertex exactly
once.
Eulerian Path Problem Given an undirected graph,
find a walk visiting every edge exactly
once. Notice that in a walk some vertices may
have been visited more than once.
The Eulerian Path problem is polynomial time
solvable. A graph has an Eulerian path if and
only if every vertex has an even degree.
The Hamiltonian Path problem is NP-complete.
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The Hamiltonian Path Problem
Application Seating assignment problem Given n
persons and a big round table of n seats, each
pair of persons may like each other or hate each
other. Can we find a seating assignment so that
everyone likes his/her neighbors (only two
neighbors)?
Construct a graph as follows. For each person,
create a vertex. For each pair of vertices, add
an edge if and only if they like each other.
Then the problem of finding a seating assignment
is reduced to the Hamiltonian path problem.
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The Traveling Salesman Problem
Traveling Salesman Problem (TSP) Given a
complete graph with nonnegative edge costs, Find
a minimum cost cycle visiting every vertex
exactly once.
Given a number of cities and the costs of
traveling from any city to any other city, what
is the cheapest round-trip route that visits each
city exactly once and then returns to the
starting city?
One of the most well-studied problem in
combinatorial optimization.
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NP-completeness of Traveling Salesman Problem
Traveling Salesman Problem (TSP) Given a
complete graph with nonnegative edge costs, Find
a minimum cost cycle visiting every vertex
exactly once.
Hamiltonian Path Problem Given an undirected
graph, find a cycle visiting every vertex exactly
once.
The Hamiltonian path problem is a special case of
the Traveling Salesman Problem.

• For each edge, we add an edge of cost 1.
• For each non-edge, we add an edge of cost 2.

• If there is a Hamiltonian path, then there is a
  cycle of cost n.
• If there is no Hamiltonian path, then every
  cycle has cost at least n1.

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Inapproximability of Traveling Salesman Problem
Theorem There is no constant factor
approximation algorithm for TSP,
unless PNP.
Idea Use the Hamiltonian path problem.

• For each edge, we add an edge of cost 1.
• For each non-edge, we add an edge of cost 2

nk.

• If there is a Hamiltonian path, then there is a
  cycle of cost n.
• If there is no Hamiltonian path, then every
  cycle has cost greater than nk.

• So, if you have a k-approximation algorithm for
  TSP,
• one just needs to check if the returned solution
  is at most nk.
• If yes, then the original graph has a
  Hamiltonian path.
• Otherwise, the original graph has no Hamiltonian
  path.

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Inapproximability of Traveling Salesman Problem
Theorem There is no constant factor
approximation algorithm for TSP,
unless PNP.
This type of theorem is called hardness result
in the literature. Just like their names, usually
they are very hard to obtain.

• If there is a Hamiltonian path, then there is a
  cycle of cost n.
• If there is no Hamiltonian path, then every
  cycle has cost greater than nk.

The strategy is usually like this. This creates a
gap between yes and no instances. The bigger the
gap, the problem is harder to approximate.
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Approximation Algorithm for TSP
There is no constant factor approximation
algorithm for TSP, what can we do then?
Observation in real world situation, the costs
satisfy triangle inequality.
a
a b c
c
For example, think of cost of an edge as the
distance between two points.
b
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Approximation Algorithm for Metric TSP
Metric Traveling Salesman Problem (metric
TSP) Given a complete graph with edge costs
satisfying triangle inequalities, Find a minimum
cost cycle visiting every vertex exactly once.
1
First check the bad example.
nk
This violates triangle inequality!
1
How could triangle inequalities help in finding
approximation algorithm?
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Lower Bounds for TSP
What can be a good lower bound to the cost of TSP?
A tour contains a matching.
Let OPT be the cost of an optimal tour, since a
tour contains two matchings, the cost of a
minimum weight perfect matching is at most OPT/2.
A tour contains a spanning tree.
So, the cost of a minimum spanning tree is at
most OPT.
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Spanning Tree and TSP
Let the thick edges have cost 1, And all other
edges have cost greater than 1.
So the thick edges form a minimum spanning tree.
But it doesnt look like a Hamiltonian cycle at
all!
Edge cost at most 2
Consider a Hamiltonian cycle. The costs of the
edges which are not in the minimum spanning tree
might have very high costs.
Not really! Each such edge has cost at most
2 because of the triangle inequality.
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Spanning Tree and TSP
a
e
b
d
c
Strategy Construct the TSP tour from a minimum
spanning tree.
Use the edges in the minimum spanning tree as
many as possible.
The center vertex already has degree 2, skip to
the next vertex.
Cost of a minimum spanning tree
Cost e a
a b
b c
c d
d e
2a 2b 2c 2d 2e 2(a b c d e)
MST OPT
SOL 2MST

SOL 2OPT
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Spanning Tree and TSP
How to formalize the idea of following a
minimum spanning tree?
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Spanning Tree and TSP
How to formalize the idea of following a
minimum spanning tree?
Key idea double all the edges and find an
Eulerian tour.
This graph has cost 2MST.
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Spanning Tree and TSP
How to formalize the idea of following a
minimum spanning tree?
Key idea double all the edges and find an
Eulerian tour.
This graph has cost 2MST.
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Spanning Tree and TSP
Strategy shortcut this Eulerian tour.
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Spanning Tree and TSP
By triangle inequalites, the shortcut tour is not
longer than the Eulerian tour.
Each directed edge is used exactly once in the
shortcut tour.
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A 2-Approximation Algorithm for Metric TSP

• (Metric TSP Factor 2)
• Find an MST, T, of G.
• Double every edge of the MST to obtain an
  Eulerian graph.
• Find an Eulerian tour, T, on this graph.
• Output the tour that visits vertices of G in the
  order of
• their first appearance in T. Let C be this
  tour.
• (That is, shortcut T)

Analysis

1. cost(T) OPT (because MST is a lower
   bound of TSP)
2. cost(T) 2cost(T) (because every edge
   appears twice)
3. cost(C) cost(T) (because of triangle
   inequalities, shortcutting)
4. So, cost(C) 2OPT

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Better approximation?
There is a 1.5 approximation algorithm for metric
TSP.
Hint use a minimum spanning tree and a maximum
matching (instead of double a minimum
spanning tree).
Major open problem Improve this to 4/3?
An aside hardness result is not an excuse to
stop working, but to guide us to
identify interesting cases.
Just for fun can we design an approximation
algorithm for the seating assignment problem?
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Minimum Steiner Tree
The Minimum Steiner Tree Problem Given an
undirected graph G with nonnegative edge costs
and whose vertices are partitioned into two sets,
required vertices and Steiner vertices, find a
minimum cost tree in G that contain all the
required vertices and any subset of the Steiner
vertices.
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Applications
Computer networks
VLSI circuit design
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Special Cases
Only two required vertices, just a shortest path.
Every vertex is required, just a minimum spanning
tree.
The general problem is NP-complete.
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Useless Edges
We must use Steiner vertices in order to
construct a good solution, because edges
connecting required vertices could be very
expensive (or even dont exist).
1
2
2
10
10
1
1
10
2
Would I ever use an edge of weight 10 (an
expensive edge)?
The purpose of using such an edge is to connect
two required vertices, but we can always use a
path of length 2 (which passes through the
Steiner vertices and has a total cost of 2) for
the same purpose!
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Metric Closure
1
2
2
10
10
1
1
10
2
In general, we can always replace the cost of an
edge by the cost of a cheapest (shortest) path.
We call the resulting graph the metric closure of
the original graph.
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Metric Closure
Claim. In the metric closure, the edge costs
satisfy triangle inequalities.
v
(Proof by contradiction) Suppose c(vw) lt c(uv)
c(uw). There is a path P(uv) between u and v
with cost c(uv), and a path P(uw) between u and w
with cost c(uw). The union of P(uv) and P(uw)
contains a path P from v to w with cost at most
c(uv) c(uw). This contradicts that c(vw)
represents the cost of a shortest path from v to
w (since P is shorter).
c(uv)
c(vw)
u
c(uw)
w
Note that the metric closure is a complete graph.
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Metric Closure
Lemma The cost of a minimum Steiner tree in a
graph G the cost of a minimum
Steiner tree in its metric closure.
First The cost of a minimum Steiner tree in a
graph G the cost of a minimum
Steiner tree in its metric closure.
Just take the same tree in the metric closure
in G
(Proof) Let T be a minimum Steiner tree in G.
For each edge uv, The cost of uv in the metric
closure is at most the cost of uv in G. So, the
same T in the metric closure is no more expensive.
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Metric Closure
Next The cost of a minimum Steiner tree in a
graph G the cost of a minimum
Steiner tree in its metric closure.
in G
in metric closure
take the union of the shortest paths
(Proof) Let T be a minimum Steiner tree in the
metric closure. Consider the union U of the
shortest paths in the original graph G. The union
U has cost at most the cost of T. Clearly, U
connects all the required vertices. So, U
contains a Steiner tree T. Hence, T has cost
at most the cost of T.
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Metric Steiner Tree
Lemma The cost of a minimum Steiner tree in a
graph G the cost of a minimum
Steiner tree in its metric closure.
Claim. In the metric closure, the edge costs
satisfy triangle inequalities.
Without loss of generality, we can consider the
metric TSP problem.
Metric Steiner Tree Problem Given a complete
graph with edge costs satisfying triangle
inequalities, find a minimum Steiner tree (which
connect all the required vertices).
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Metric Steiner Tree
A Steiner tree connects all the required vertices.
1
2
2
10
10
Can we just find a minimum spanning
tree connecting all the required vertices?
1
1
10
2
In other words, forget about all Steiner vertices!
How bad can it be?
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Spanning Tree and Steiner Tree
a
e
b
d
c
Cost of a minimum Steiner tree
MST a b
b c
c d
d e
2a 2b 2c 2d 2e 2(a b c d e)
MST 2OPT
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Spanning Tree and Steiner Tree
How to formalize the idea?
Let the cost of this Steiner tree be OPT.
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Spanning Tree and Steiner Tree
How to formalize the idea?
Key idea double all the edges and find an
Eulerian tour.
This graph has cost 2OPT.
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Spanning Tree and Steiner Tree
How to formalize the idea?
Key idea double all the edges and find an
Eulerian tour.
This graph has cost 2OPT.
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Spanning Tree and Steiner Tree
Strategy shortcut this Eulerian tour.
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Spanning Tree and Steiner Tree
By triangle inequalites, the shortcut tour is not
longer than the Eulerian tour.
So, the cost of MST the cost of Eulerian graph
2OPT.
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A 2-Approximation for Metric Steiner Tree

• (Metric Steiner Tree Factor 2)
• Compute the metric closure G of G.
• Find a minimum spanning tree T of G.
• Construct the union U of the shortest paths in G
  which
• correspond to edges in T.
• 4. Output the Steiner tree T contained in U.

Analysis

1. cost(T) cost(U) (because T is contained
   in U)
2. cost(U) cost(T) (because G is the metric
   closure of G)
3. cost(T) 2OPT (because of triangle
   inequalities, shortcutting)
4. So, cost(T) 2OPT

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Better approximation?
(Robin Zelikovsky) There is a 1.55 approximation
algorithm for minimum Steiner tree.

• Just for fun 1 Show that the 2-approximation
  algorithm is equivalent to the following
  algorithm
• Find a cheapest path that connects two required
  vertices which are not yet connected.
• - Repeat until all required vertices are
  connected.

Just for fun 2 Design a polynomial time
algorithm to find a minimum Steiner tree for k
required vertices, when k is fixed.